   # A 2.00−L sample of O 2 ( g ) was collected over water at a total pressure of 785 torr and 25°C. When the O 2 ( g ) was dried (water vapor removed), the gas had a volume of 1.94 L at 25°C and 785 torr. Calculate the vapor pressure of water at 25°C. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 8, Problem 133AE
Textbook Problem
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## A 2.00−L sample of O2(g) was collected over water at a total pressure of 785 torr and 25°C. When the O2(g) was dried (water vapor removed), the gas had a volume of 1.94 L at 25°C and 785 torr. Calculate the vapor pressure of water at 25°C.

Interpretation Introduction

Interpretation: For a given data, vapor pressure of water should be determined.

Concept introduction:

Boyle’s law can defined as,

A gas kept in the particular container at constant temperature, the excreted pressure is inversely proportional to the volume of the container. That is the volume of a container going to decreases by increasing the applied pressure on the gas container. It can be explained as when the pressure of a gas container increases, the distance between the gaseous particles in the unit area is decreases and it becomes denser.  By increasing the gaseous particles in the unit area of a container the volume of a container also decreases.

Mathematically,

P1V

or

PV=K

Where,

P = pressure in atmospheres

V= volumes in liters

K= a constant for a particular gas at given temperature

By comparing two systems at different conditions

P1V1=P2V2

By rearranging the equation,

P1=P2V2V1

### Explanation of Solution

Explanation

• To determine: Vapor pressure of oxygen

Vapor pressure of oxygen, P1=761torr

According to Boyle’s law,

P1V1=P2V2

By rearranging the equation,

P1=P2V2V1

Herethegivendata'sare,P2=785torrV2=1.94LV1=2.00LP1=785torr×1

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